:PROPERTIES:
:header-args: :noweb-ref sub-nat-nums
:ID:       little-schemer-chapter-four-part-2
:CREATED: [2020-12-02 Wed 20:46]
:END:
:LOGBOOK:
- State "DONE"       from "TODO"       [2020-12-02 Wed 20:57]
:END:

How do we define subtraction in terms of =sub1=? Based on the content of this book so far, it's likely got something to do with recursion.

Again, going from the solution to =o+= we can guess that we'll be asking a series of questions with =cond=, and that we'll start by checking for 0. We know that there'll be 2 arguments, =n= and =m=. We'll need to subtract =m= from =n=, and we'll need to stop when one of them hits =0=. This kinda tells us that both arguments are gonna need to be subtracted.

If we were using real numbers, I'd be worried here because we only have =sub1= to work with, and negative numbers make that complex - but with natural numbers, subtracting both works just fine. We'll never have to deal with =6 - -3= or whatever.

In essence, what we need to do here is =(sub1 n)= =m= times. Let's give that a go:

#+NAME: o-
#+BEGIN_SRC racket :lang racket
(define o-
  (lambda (n m)
    (cond
      ((zero? m) n)
      (else (o- (sub1 n) (sub1 m))))))


(o- 14 3)
(o- 17 9)
#+END_SRC

#+RESULTS: o-
: 11
: 8

That seems to work pretty well. Looking at the definition in the book, there's the same pattern again - they define =o-= in the same way that I have, but the last line is

: (else (sub1 (o- n (sub1 m))))

I'm a little concerned that I'm missing some point or other here, but it could be just that I think in a different way. I'm gonna leave it, and hope that it doesn't come back to bite me!